# d orbital irreducible representations in metal in ML4 model [closed]

In ML4 (Metal—4 ligands)model which has a square planar shape and D4h point group, what is the irreducible representation for the 5 kinds of d orbitals in the central metal?

• You will need to place x,y,z vectors on each atom and work out what happens to them as you perform each of the operations in the point group. Count 0 if the atom moves and $\pm 1$ if it unchanged or reversed. Most book on symmetry go through examples. Commented Jun 10, 2021 at 6:09
• I almost know the procedure but probably wrong with first finding out the reducible representations because I cannot get an integer for the number of A1g or B1g in the later step. Commented Jun 10, 2021 at 7:28
• @user111996 Please take a site tour and review our policies for asking questions and including your efforts within. Commented Jun 10, 2021 at 17:56

I had this example worked out. You should be able to copy the method, it is a bit long but easier than using the equations as errors are easily made there.

To find the symmetry of the vibrations of a C2v molecule such as H2O or SO2. (Note that the method only works this way for 180 deg rotation groups if not you need to use cos/ sine to find amounts.)

The method is:

(1) Place three (unit) vectors on each atom and along x, y, z axes.

(2) Move each vector in turn and according to each of the symmetry operations; in C2v these are: identity E, rotation about the principal axis C2, and two reflections $$\sigma_{xz}$$ and $$\sigma_{zy}$$.

(3) Add up the number of vectors unmoved, moved or inverted and this generates our reducible representation which we label $$\Gamma_R$$. We then have to use this to produce an irreducible representation

(4) if a vector is unmoved we add 1, if it is inverted we add -1, if it is moved we add 0.

(5) Place the total count under each column of the point group table operators and this is the $$\Gamma_R$$.

example

Rotation C2: The C2 axis lies along z1 therefore z1 remains unchanged so we count $$z_1 \equiv 1$$, $$y_1$$ and $$x_1$$ are both inverted by the $$180^\text{o}$$ rotation so $$y_1 \equiv -1,\; x_1 \equiv -1$$. Each vector on the H atoms is moved to the opposite side, so exchange places, and as they have completely moved they each count 0.

So the total for $$C_2$$ rotation is $$+1-1 -1=-1$$.

The reduced representation looks like

$$\displaystyle \begin{array}\\ C_{2v} & E & C_2 & \sigma_{xz} & \sigma_{zy}\\ \hline \Gamma_R & 9 & -1 & 3 & 1\\ \hline \end{array}$$

Now we need to reduce this representation into its irreducible representations present in the character table. Build a new table;

(1) add the classes on the top line E, C2 etc.

(2) The next row is the reducible representation $$\Gamma_R$$ plus S as the sum of the characters in that row and also S/h as the result we want which is the number of times that representation is present in $$\Gamma_R$$.

( $$h$$ is the order of the group which for C2v = 4; its different for other groups, you will need to calculate it from the point group table. The order is the number of columns times the number in each class . This is 1 for each of the operations in C2v )

(3) subsequent rows start with each Mulliken symbol in the Character table

To calculate the irreducible representation start each row with the Mulliken labels (A1, A2 etc) from the point group (character) table then for each entry in the table:

Multiply the number in the class (1 for C2 and each of the $$\sigma$$'s) by the number under it in GR and by the entry in the point group table: e.g.

$$\text{( Number in class)} \times \text{( value in } \Gamma_R) \times \text{( entry in character table)}$$

for the $$A_2-\sigma_{xz}$$ entry the product is $$(1)x(3)x(-1)=-3$$. The $$-1$$ comes from the point group table entry at position $$A_2 : \sigma_{xz}$$

The reduced representation is therefore formed from $$3A_1, 1A_2, 3B_1$$ and $$2B_2$$ irreducible representations;

$$\Gamma_R=3A_1 + A_2+3B_1+2B_2$$

Because we have used (unit) vectors on each atom we have 3N coordinates, 3 are from whole molecule translations and 3 from rotations leaving 3N-6 vibrations.

The translations are in each of the x, y, and z directions and also rotations about x, y and x axes and these have to be removed to reveal the 3N-6 vibrations.

The translations and rotations are easily found as they are described by the irreducible representations in the point group table under x, y, z and Rx, Ry and Rz.

The values for C2v are:

Translations $$A_1 + B_1 + B_2$$ Rotations $$A_2 + B_1 + B_2$$

Total trans+rot $$A_1 +A_2 +2B_1 +2B_2$$

If we subtract these from the total $$\Gamma_R = 3A_1 + A_2 + 3B_1 + 2B_2$$ we have calculated we obtain the symmetry of the remaining vibrations:

Vibrations $$= 2A_1 + B_1$$.